Problem: Simplify the following expression: $y = \dfrac{2x^2+9x- 5}{x + 5}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(2)}{(-5)} &=& -10 \\ {a} + {b} &=& &=& {9} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-10$ and add them together. Remember, since $-10$ is negative, one of the factors must be negative. The factors that add up to ${9}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-1}$ and ${b}$ is ${10}$ $ \begin{eqnarray} {ab} &=& ({-1})({10}) &=& -10 \\ {a} + {b} &=& {-1} + {10} &=& 9 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({2}x^2 {-1}x) + ({10}x {-5}) $ Factor out the common factors: $ x(2x - 1) + 5(2x - 1)$ Now factor out $(2x - 1)$ $ (2x - 1)(x + 5)$ The original expression can therefore be written: $ \dfrac{(2x - 1)(x + 5)}{x + 5}$ We are dividing by $x + 5$ , so $x + 5 \neq 0$ Therefore, $x \neq -5$ This leaves us with $2x - 1; x \neq -5$.